False they don't have to be whole numbers. As long as they're greater than 0
Answer:
(-5, -30)
Step-by-step explanation:
We start with the equation:
y = 6x
y = -4x - 50
Since both of them are equal to y, we can substitute the 6x into y. So:
6x = -4x - 50
Now we can solve for x. Add 4x to both sides...
6x = -4x - 50
+4x +4x
10x = -50
Now, divide both sides by 10.
<u>10x</u> = <u>-50</u>
10 10
x = -5
So, we now know that x is -5. Now we know this, we can substitute -5 where we see x. It works in both equations, so I will do both.
y = 6(-5)
y = -30
And...
y = -4(-5) - 50
y = 20 - 50
y = -30
As you can see, both equations show that y = -30. To check our work, let's plug in the values.
-30 = 6(-5)
-30 = -30
Also:
-30 = -4(-5) - 50
-30 = -20 - 50
-30 = -30
So, we can see that both of the equations gave us the same answer. So, our answer is x = -5 and y = -30, so the ordered pair is: (-5, -30). Hope this helps you!
Answer:
= -2x
Step-by-step explanation:
3x and -5x are like terms and -10 and 10 are also like terms. like terms can be added together
= -2x cause the 10's cancel out
Two main facts are needed here:
1. The logarithm 
, regardless of the base of the logarithm, exists for 
.
2. The square root 
exists for 
.
(in both cases we're assuming real-valued functions only)
By (2) we know that 
exists if 
, or 
.
By (1), we know that 
exists if 
, or 
. But as long as the square root exists, it will always be positive, so this condition will always be met.
Ultimately, then, we only require , so the function has domain 
.
To determine the range, we need to know that, in their respective domains, and increase monotonically without bound. We also know that 
at minimum, at which point the square root term vanishes, so the least value the function takes on is 
. Then its range would be 
.
